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A bar of iron is 10 cm at `20^(@)C`. At `19^(@)C` it will be (`alpha` of iron `=11xx10.//^(@)C`)A. `11xx10^(-6)` cm longerB. `11xx10^(-6)` cm shorterC. `11xx10^(-5)` cm shorteD. `11xx10^(-5)` cm longer |
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Answer» Correct Answer - C Chane is length, `Deltal=l_(o)alphaDeltaT=10xx11xx10^(-6)(19-20)` `" "=-11xx10^(-5)` cm |
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